如何使用 Python 组合嵌套列表中的每两个值？答案

【问题标题】：How to combine every two values in a nested list using Python?如何使用 Python 组合嵌套列表中的每两个值？
【发布时间】：2019-11-04 20:34:07
【问题描述】：

nested_lst = [['AAA, CEO','BBB, Global Head','CCC, Regional Manager','DDD, Analyst','People also report to CCC','XXX, Analyst','VVV, Analyst'],
['AAA, CEO','BBB, Global Head','EEE, Regional Manager','FFF, Analyst','People also report to EEE','SSS, Analyst','LLL, Analyst'],
['AAA, CEO','BBB, Global Head','PPP, Regional Manager','MMM, Manager','People report to MMM','GGG, Associate','People also report to EEE','ZZZ, Junior Analyst','UUU, Contractor']]

大家好，我有一个上面显示的嵌套列表，我想组合该嵌套列表中的每两个元素。期望的输出是：

[['AAA, CEO & BBB, Global Head','BBB, Global Head & CCC, Regional Manager','CCC, Regional Manager & DDD, Analyst','DDD, Analyst & People also report to CCC',...],
['AAA, CEO & BBB, Global Head','BBB, Global Head & EEE, Regional Manager','EEE, Regional Manager & FFF, Analyst',...],
['AAA, CEO & BBB, Global Head','BBB, Global Head & PPP, Regional Manager',...]]

我尝试了[' & '.join(x) for x in zip(nested_lst[2][0::2], nested_lst[2][1::2]) ]，但它不适用于嵌套列表并且没有组合元素。

谁能帮我解答这个问题？感谢您的帮助！

【问题讨论】：

标签： python list list-comprehension nested-lists

【解决方案1】：

你可以使用itertools.pairwise配方：

result = [list(map(' & '.join, pairwise(l))) for l in nested_lst]

或者没有 itertools：

result = [list(map(' & '.join, zip(l, l[1:]))) for l in nested_lst]

结果：

[['AAA, CEO & BBB, Global Head', 'CCC, Regional Manager & DDD, Analyst', 'People also report to CCC & XXX, Analyst', 'VVV,Analyst'], ['AAA, CEO & BBB, Global Head', 'EEE, Regional Manager & FFF, Analyst', 'People also report to EEE & SSS, Analyst', 'LLL, Analyst'], ['AAA, CEO & BBB, Global Head', 'PPP, Regional Manager & MMM, Manager', 'People report to MMM & GGG, Associate', 'People also report to EEE & ZZZ, Junior Analyst', 'UUU, Contractor']]

【讨论】：

谢谢贾布。 itertools.pairwise 看起来很棒！

【解决方案2】：

您的列表组合已接近。它需要一个嵌套循环和一些更改。

list_comp：

[[f'{x} & {y}' for x, y in zip(lst, lst[1:])] for lst in nested_lst]

>>> output



[['AAA, CEO & BBB, Global Head',
 'BBB, Global Head & CCC, Regional Manager',
  'CCC, Regional Manager & DDD, Analyst',
  'DDD, Analyst & People also report to CCC',
  'People also report to CCC & XXX, Analyst',
  'XXX, Analyst & VVV, Analyst'],
 ['AAA, CEO & BBB, Global Head',
  'BBB, Global Head & EEE, Regional Manager',
  'EEE, Regional Manager & FFF, Analyst',
  'FFF, Analyst & People also report to EEE',
  'People also report to EEE & SSS, Analyst',
  'SSS, Analyst & LLL, Analyst'],
 ['AAA, CEO & BBB, Global Head',
  'BBB, Global Head & PPP, Regional Manager',
  'PPP, Regional Manager & MMM, Manager',
  'MMM, Manager & People report to MMM',
  'People report to MMM & GGG, Associate',
  'GGG, Associate & People also report to EEE',
  'People also report to EEE & ZZZ, Junior Analyst',
  'ZZZ, Junior Analyst & UUU, Contractor']]

【讨论】：

谢谢你，布赖恩。您的回答非常有帮助！